The given expression is the limit as $x \to 0$ of $\frac{\sin(x)}{\sqrt{\log(1+x)}}$:

$\lim_{x \to 0} \frac{\sin(x)}{\sqrt{\log(1+x)}}$

To solve this, let's first analyze the behavior of both the numerator and the denominator as $x \to 0$.

1. Numerator: As $x \to 0$, \sin(x) \approx x \quad \text{(since \sin(x)$behaves like$x near zero)}. So, $\sin(x) \to 0$ as $x \to 0$.

2. Denominator: For small values of $x$, we can approximate $\log(1+x)$ using the first term in its Taylor expansion: \log(1+x) \approx x \quad \text{(since \log(1+x) \approx x$for small$x)}. Hence, $\sqrt{\log(1+x)} \approx \sqrt{x}.$

Now the expression becomes:

$\lim_{x \to 0} \frac{x}{\sqrt{x}} = \lim_{x \to 0} \sqrt{x}.$

As $x \to 0$, $\sqrt{x} \to 0$.

Thus, the limit of the given expression as $x \to 0$ is:

$\boxed{0}.$

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. What is the Taylor expansion of $\sin(x)$?
2. How can L'Hopital's rule be applied to limits like this one?
3. What are the general properties of logarithmic functions near zero?
4. Can you approximate higher-order terms in the expansion of $\log(1+x)$?
5. What is the limit of $\frac{\sin(x)}{\log(1+x)}$ as $x \to 0$?

Tip: In many limits involving trigonometric and logarithmic functions, approximations using Taylor series are useful to simplify the expressions.